# Graph Potential Energy over Position

1. Dec 4, 2009

### merzperson

1. The problem statement, all variables and given/known data

The figure shows the force Fx exerted on a particle that moves along the x-axis. Draw a graph of the particle's potential energy as a function of position x from 0m to 1.1m. Let U be zero at x=0m.

2. Relevant equations

W = F*d
U = mgh

3. The attempt at a solution

I'm pretty confused about how to start this problem. How do you know what equation to use in this situation?

My best idea is to use W = F*d because it includes both Force and Displacement, and relates them with Work, which is equal to the change in Energy of the system. However, I don't know how to use this to find the Potential Energy at a given Position. Please help!

2. Dec 4, 2009

### nasu

You need the general relationship between potential energy and force.
U=mgh is just a special case and it has nothing to do with this problem.

3. Dec 4, 2009

### Fightfish

Have you seen the relation $$F = - \frac{dU}{dr}$$ before?
If not, you probably have to review your textbook or lecture notes.
W = F.d only holds true for a constant force, which it clearly is not in this case.

4. Dec 4, 2009

### merzperson

Thanks guys!

Using the equation F = -dU/dr we can solve for dU:

dU = -Fdr

And then plug in the values for force (F) and displacement (dr). However, what do I plug in for the displacement? For example, at x=0.5

dU = -(2)(0.5) = -1

Is this correct? I'm not sure I'm using the displacement correctly. Also, once I find dU, how do I find U? Do I just take a small interval? Thank you for your help.

5. Dec 4, 2009

### nasu

In order to find U you need to integrate the force over the position.
You can find F(x) from the graph (separately for the two sections) and then integrate, again for the two sections.

6. Dec 5, 2009

### Fightfish

Hm...actually on second thought, I'm dubious as to whether the potential energy of the system actually changes. Assuming that we are not doing any work against forces like gravity, since the force acts along the direction of motion of the particle, the work done on the body goes solely to increasing its kinetic energy.

7. Dec 5, 2009

### nasu

Of course there is a change in potential energy. Once again, gravitational potential energy is just one special case of potential energy.
The change in potential energy is equal to the the work done, with a minus sign.
The change in kinetic energy is equal to the work.
If the KE increases, the PE decreases.
If you need a special case to understand better, the force pictured here behaves like an elastic force (for each one of the two segments, not for the whole thing). Think about a ball attached to a spring and how the potential and kinetic energies vary during the motion.

8. Dec 5, 2009

### Fightfish

That I do not agree with. That the change in KE is equivalent to the change in PE magnitude-wise is only true for internal forces acting within the system of consideration. For a gravitational system of masses, or a spring-mass system, that clearly holds true. But what of situations such as a force pushing a block horizontally along the ground, or compression of a gas?

9. Dec 5, 2009

### merzperson

Thanks for all your help, I solved the problem. The reason for my confusion was because I never learned to integrate. I was doing the problem backwards.

10. Dec 5, 2009

### nasu

If the force is conservative then it can be derived from a potential. The problem asks to calculate the potential energy so I assume this is a force that can be derived from a potential.
I mean for the problem at hand.

I did not mean and did not make a general statement. The kinetic/potential energy transformation is only valid for these cases when the potential energy can be defined and I think is reasonable to assume that this is the case in this problem.

It is interesting to think about the case of an object pulled along a horizontal surface, and accelerated.
Where does the kinetic energy comes from? It depends on where that force comes from but somewhere some other kind of energy must decrease, right?